Trigonometry

These formulas come from the definition, Euler's formula ${\displaystyle e^{ix}=\cos x+i\sin x,}$ and elementary trigonometric identities.

${\displaystyle {\begin{aligned}c_{1}(n)&=1\\c_{2}(n)&=\cos n\pi \\c_{3}(n)&=2\cos {\tfrac {2}{3}}n\pi \\c_{4}(n)&=2\cos {\tfrac {1}{2}}n\pi \\c_{5}(n)&=2\cos {\tfrac {2}{5}}n\pi +2\cos {\tfrac {4}{5}}n\pi \\c_{6}(n)&=2\cos {\tfrac {1}{3}}n\pi \\c_{7}(n)&=2\cos {\tfrac {2}{7}}n\pi +2\cos {\tfrac {4}{7}}n\pi +2\cos {\tfrac {6}{7}}n\pi \\c_{8}(n)&=2\cos {\tfrac {1}{4}}n\pi +2\cos {\tfrac {3}{4}}n\pi \\c_{9}(n)&=2\cos {\tfrac {2}{9}}n\pi +2\cos {\tfrac {4}{9}}n\pi +2\cos {\tfrac {8}{9}}n\pi \\c_{10}(n)&=2\cos {\tfrac {1}{5}}n\pi +2\cos {\tfrac {3}{5}}n\pi \\\end{aligned}}}$

and so on (OEIS: A000012, OEIS: A033999, OEIS: A099837, OEIS: A176742,.., OEIS: A100051,...). c_q(n) is always an integer.

Kluyver

Let ${\displaystyle \zeta _{q}=e^{\frac {2\pi i}{q}}.}$ Then ζ_q is a root of the equation x^q − 1 = 0. Each of its powers,

${\displaystyle \zeta _{q},\zeta _{q}^{2},\ldots ,\zeta _{q}^{q-1},\zeta _{q}^{q}=\zeta _{q}^{0}=1}$

is also a root. Therefore, since there are q of them, they are all of the roots. The numbers ${\displaystyle \zeta _{q}^{n}}$ where 1 ≤ n ≤ q are called the q-th roots of unity. ζ_q is called a primitive q-th root of unity because the smallest value of n that makes ${\displaystyle \zeta _{q}^{n}=1}$ is q. The other primitive q-th roots of unity are the numbers ${\displaystyle \zeta _{q}^{a}}$ where (a, q) = 1. Therefore, there are φ(q) primitive q-th roots of unity.

Thus, the Ramanujan sum c_q(n) is the sum of the n-th powers of the primitive q-th roots of unity.

It is a fact^[3] that the powers of ζ_q are precisely the primitive roots for all the divisors of q.

Example. Let q = 12. Then

${\displaystyle \zeta _{12},\zeta _{12}^{5},\zeta _{12}^{7},}$ and ${\displaystyle \zeta _{12}^{11}}$ are the primitive twelfth roots of unity,

${\displaystyle \zeta _{12}^{2}}$ and ${\displaystyle \zeta _{12}^{10}}$ are the primitive sixth roots of unity,

${\displaystyle \zeta _{12}^{3}=i}$ and ${\displaystyle \zeta _{12}^{9}=-i}$ are the primitive fourth roots of unity,

${\displaystyle \zeta _{12}^{4}}$ and ${\displaystyle \zeta _{12}^{8}}$ are the primitive third roots of unity,

${\displaystyle \zeta _{12}^{6}=-1}$ is the primitive second root of unity, and

${\displaystyle \zeta _{12}^{12}=1}$ is the primitive first root of unity.

Therefore, if

${\displaystyle \eta _{q}(n)=\sum _{k=1}^{q}\zeta _{q}^{kn}}$

is the sum of the n-th powers of all the roots, primitive and imprimitive,

${\displaystyle \eta _{q}(n)=\sum _{d\mid q}c_{d}(n),}$

and by Möbius inversion,

${\displaystyle c_{q}(n)=\sum _{d\mid q}\mu \left({\frac {q}{d}}\right)\eta _{d}(n).}$

It follows from the identity x^q − 1 = (x − 1)(x^q−1 + x^q−2 + ... + x + 1) that

${\displaystyle \eta _{q}(n)={\begin{cases}0&q\nmid n\\q&q\mid n\\\end{cases}}}$

and this leads to the formula

${\displaystyle c_{q}(n)=\sum _{d\mid (q,n)}\mu \left({\frac {q}{d}}\right)d,}$

published by Kluyver in 1906.^[4]

This shows that c_q(n) is always an integer. Compare it with the formula

${\displaystyle \phi (q)=\sum _{d\mid q}\mu \left({\frac {q}{d}}\right)d.}$

von Sterneck

It is easily shown from the definition that c_q(n) is multiplicative when considered as a function of q for a fixed value of n:^[5] i.e.

${\displaystyle {\mbox{If }}\;(q,r)=1\;{\mbox{ then }}\;c_{q}(n)c_{r}(n)=c_{qr}(n).}$

From the definition (or Kluyver's formula) it is straightforward to prove that, if p is a prime number,

${\displaystyle c_{p}(n)={\begin{cases}-1&{\mbox{ if }}p\nmid n\\\phi (p)&{\mbox{ if }}p\mid n\\\end{cases}},}$

and if p^k is a prime power where k > 1,

${\displaystyle c_{p^{k}}(n)={\begin{cases}0&{\mbox{ if }}p^{k-1}\nmid n\\-p^{k-1}&{\mbox{ if }}p^{k-1}\mid n{\mbox{ and }}p^{k}\nmid n\\\phi (p^{k})&{\mbox{ if }}p^{k}\mid n\\\end{cases}}.}$

This result and the multiplicative property can be used to prove

${\displaystyle c_{q}(n)=\mu \left({\frac {q}{(q,n)}}\right){\frac {\phi (q)}{\phi \left({\frac {q}{(q,n)}}\right)}}.}$

This is called von Sterneck's arithmetic function.^[6] The equivalence of it and Ramanujan's sum is due to Hölder.^[7][8]

Other properties of c_q(n)

For all positive integers q,

${\displaystyle {\begin{aligned}c_{1}(q)&=1\\c_{q}(1)&=\mu (q)\\c_{q}(q)&=\phi (q)\\c_{q}(m)&=c_{q}(n)&&{\text{for }}m\equiv n{\pmod {q}}\\\end{aligned}}}$

For a fixed value of q the absolute value of the sequence ${\displaystyle \{c_{q}(1),c_{q}(2),\ldots \}}$ is bounded by φ(q), and for a fixed value of n the absolute value of the sequence ${\displaystyle \{c_{1}(n),c_{2}(n),\ldots \}}$ is bounded by n.

If q > 1

${\displaystyle \sum _{n=a}^{a+q-1}c_{q}(n)=0.}$

Let m₁, m₂ > 0, m = lcm(m₁, m₂). Then^[9] Ramanujan's sums satisfy an orthogonality property:

${\displaystyle {\frac {1}{m}}\sum _{k=1}^{m}c_{m_{1}}(k)c_{m_{2}}(k)={\begin{cases}\phi (m)&m_{1}=m_{2}=m,\\0&{\text{otherwise}}\end{cases}}}$

Let n, k > 0. Then^[10]

${\displaystyle \sum _{\stackrel {d\mid n}{\gcd(d,k)=1}}d\;{\frac {\mu ({\tfrac {n}{d}})}{\phi (d)}}={\frac {\mu (n)c_{n}(k)}{\phi (n)}},}$

known as the Brauer - Rademacher identity.

If n > 0 and a is any integer, we also have^[11]

${\displaystyle \sum _{\stackrel {1\leq k\leq n}{\gcd(k,n)=1}}c_{n}(k-a)=\mu (n)c_{n}(a),}$

due to Cohen.

Generating functions

The generating functions of the Ramanujan sums are Dirichlet series:

${\displaystyle \zeta (s)\sum _{\delta \,\mid \,q}\mu \left({\frac {q}{\delta }}\right)\delta ^{1-s}=\sum _{n=1}^{\infty }{\frac {c_{q}(n)}{n^{s}}}}$

is a generating function for the sequence c_q(1), c_q(2), ... where q is kept constant, and

${\displaystyle {\frac {\sigma _{r-1}(n)}{n^{r-1}\zeta (r)}}=\sum _{q=1}^{\infty }{\frac {c_{q}(n)}{q^{r}}}}$

is a generating function for the sequence c₁(n), c₂(n), ... where n is kept constant.

There is also the double Dirichlet series

${\displaystyle {\frac {\zeta (s)\zeta (r+s-1)}{\zeta (r)}}=\sum _{q=1}^{\infty }\sum _{n=1}^{\infty }{\frac {c_{q}(n)}{q^{r}n^{s}}}.}$

σ_k(n)

σ_k(n) is the divisor function (i.e. the sum of the k-th powers of the divisors of n, including 1 and n). σ₀(n), the number of divisors of n, is usually written d(n) and σ₁(n), the sum of the divisors of n, is usually written σ(n).

If s > 0,

${\displaystyle {\begin{aligned}\sigma _{s}(n)&=n^{s}\zeta (s+1)\left({\frac {c_{1}(n)}{1^{s+1}}}+{\frac {c_{2}(n)}{2^{s+1}}}+{\frac {c_{3}(n)}{3^{s+1}}}+\cdots \right)\\\sigma _{-s}(n)&=\zeta (s+1)\left({\frac {c_{1}(n)}{1^{s+1}}}+{\frac {c_{2}(n)}{2^{s+1}}}+{\frac {c_{3}(n)}{3^{s+1}}}+\cdots \right)\end{aligned}}}$

Setting s = 1 gives

${\displaystyle \sigma (n)={\frac {\pi ^{2}}{6}}n\left({\frac {c_{1}(n)}{1}}+{\frac {c_{2}(n)}{4}}+{\frac {c_{3}(n)}{9}}+\cdots \right).}$

If the Riemann hypothesis is true, and ${\displaystyle -{\tfrac {1}{2}}<s<{\tfrac {1}{2}},}$

${\displaystyle \sigma _{s}(n)=\zeta (1-s)\left({\frac {c_{1}(n)}{1^{1-s}}}+{\frac {c_{2}(n)}{2^{1-s}}}+{\frac {c_{3}(n)}{3^{1-s}}}+\cdots \right)=n^{s}\zeta (1+s)\left({\frac {c_{1}(n)}{1^{1+s}}}+{\frac {c_{2}(n)}{2^{1+s}}}+{\frac {c_{3}(n)}{3^{1+s}}}+\cdots \right).}$

d(n)

d(n) = σ₀(n) is the number of divisors of n, including 1 and n itself.

${\displaystyle {\begin{aligned}-d(n)&={\frac {\log 1}{1}}c_{1}(n)+{\frac {\log 2}{2}}c_{2}(n)+{\frac {\log 3}{3}}c_{3}(n)+\cdots \\-d(n)(2\gamma +\log n)&={\frac {\log ^{2}1}{1}}c_{1}(n)+{\frac {\log ^{2}2}{2}}c_{2}(n)+{\frac {\log ^{2}3}{3}}c_{3}(n)+\cdots \end{aligned}}}$

where γ = 0.5772... is the Euler–Mascheroni constant.

φ(n)

Euler's totient function φ(n) is the number of positive integers less than n and coprime to n. Ramanujan defines a generalization of it, if

${\displaystyle n=p_{1}^{a_{1}}p_{2}^{a_{2}}p_{3}^{a_{3}}\cdots }$

is the prime factorization of n, and s is a complex number, let

${\displaystyle \varphi _{s}(n)=n^{s}(1-p_{1}^{-s})(1-p_{2}^{-s})(1-p_{3}^{-s})\cdots ,}$

so that φ₁(n) = φ(n) is Euler's function.^[17]

He proves that

${\displaystyle {\frac {\mu (n)n^{s}}{\varphi _{s}(n)\zeta (s)}}=\sum _{\nu =1}^{\infty }{\frac {\mu (n\nu )}{\nu ^{s}}}}$

and uses this to show that

${\displaystyle {\frac {\varphi _{s}(n)\zeta (s+1)}{n^{s}}}={\frac {\mu (1)c_{1}(n)}{\varphi _{s+1}(1)}}+{\frac {\mu (2)c_{2}(n)}{\varphi _{s+1}(2)}}+{\frac {\mu (3)c_{3}(n)}{\varphi _{s+1}(3)}}+\cdots .}$

Letting s = 1,

${\displaystyle \varphi (n)={\frac {6}{\pi ^{2}}}n\left(c_{1}(n)-{\frac {c_{2}(n)}{2^{2}-1}}-{\frac {c_{3}(n)}{3^{2}-1}}-{\frac {c_{5}(n)}{5^{2}-1}}+{\frac {c_{6}(n)}{(2^{2}-1)(3^{2}-1)}}-{\frac {c_{7}(n)}{7^{2}-1}}+{\frac {c_{10}(n)}{(2^{2}-1)(5^{2}-1)}}-\cdots \right).}$

Note that the constant is the inverse^[18] of the one in the formula for σ(n).

Λ(n)

Von Mangoldt's function Λ(n) = 0 unless n = p^k is a power of a prime number, in which case it is the natural logarithm log p.

${\displaystyle -\Lambda (m)=c_{m}(1)+{\frac {1}{2}}c_{m}(2)+{\frac {1}{3}}c_{m}(3)+\cdots }$

Zero

For all n > 0,

${\displaystyle 0=c_{1}(n)+{\frac {1}{2}}c_{2}(n)+{\frac {1}{3}}c_{3}(n)+\cdots .}$

This is equivalent to the prime number theorem.^[19][20]

r_2s(n) (sums of squares)

r_2s(n) is the number of way of representing n as the sum of 2s squares, counting different orders and signs as different (e.g., r₂(13) = 8, as 13 = (±2)² + (±3)² = (±3)² + (±2)².)

Ramanujan defines a function δ_2s(n) and references a paper^[21] in which he proved that r_2s(n) = δ_2s(n) for s = 1, 2, 3, and 4. For s > 4 he shows that δ_2s(n) is a good approximation to r_2s(n).

s = 1 has a special formula:

${\displaystyle \delta _{2}(n)=\pi \left({\frac {c_{1}(n)}{1}}-{\frac {c_{3}(n)}{3}}+{\frac {c_{5}(n)}{5}}-\cdots \right).}$

In the following formulas the signs repeat with a period of 4.

${\displaystyle {\begin{aligned}\delta _{2s}(n)&={\frac {\pi ^{s}n^{s-1}}{(s-1)!}}\left({\frac {c_{1}(n)}{1^{s}}}+{\frac {c_{4}(n)}{2^{s}}}+{\frac {c_{3}(n)}{3^{s}}}+{\frac {c_{8}(n)}{4^{s}}}+{\frac {c_{5}(n)}{5^{s}}}+{\frac {c_{12}(n)}{6^{s}}}+{\frac {c_{7}(n)}{7^{s}}}+{\frac {c_{16}(n)}{8^{s}}}+\cdots \right)&&s\equiv 0{\pmod {4}}\\[6pt]\delta _{2s}(n)&={\frac {\pi ^{s}n^{s-1}}{(s-1)!}}\left({\frac {c_{1}(n)}{1^{s}}}-{\frac {c_{4}(n)}{2^{s}}}+{\frac {c_{3}(n)}{3^{s}}}-{\frac {c_{8}(n)}{4^{s}}}+{\frac {c_{5}(n)}{5^{s}}}-{\frac {c_{12}(n)}{6^{s}}}+{\frac {c_{7}(n)}{7^{s}}}-{\frac {c_{16}(n)}{8^{s}}}+\cdots \right)&&s\equiv 2{\pmod {4}}\\[6pt]\delta _{2s}(n)&={\frac {\pi ^{s}n^{s-1}}{(s-1)!}}\left({\frac {c_{1}(n)}{1^{s}}}+{\frac {c_{4}(n)}{2^{s}}}-{\frac {c_{3}(n)}{3^{s}}}+{\frac {c_{8}(n)}{4^{s}}}+{\frac {c_{5}(n)}{5^{s}}}+{\frac {c_{12}(n)}{6^{s}}}-{\frac {c_{7}(n)}{7^{s}}}+{\frac {c_{16}(n)}{8^{s}}}+\cdots \right)&&s\equiv 1{\pmod {4}}{\text{ and }}s>1\\[6pt]\delta _{2s}(n)&={\frac {\pi ^{s}n^{s-1}}{(s-1)!}}\left({\frac {c_{1}(n)}{1^{s}}}-{\frac {c_{4}(n)}{2^{s}}}-{\frac {c_{3}(n)}{3^{s}}}-{\frac {c_{8}(n)}{4^{s}}}+{\frac {c_{5}(n)}{5^{s}}}-{\frac {c_{12}(n)}{6^{s}}}-{\frac {c_{7}(n)}{7^{s}}}-{\frac {c_{16}(n)}{8^{s}}}+\cdots \right)&&s\equiv 3{\pmod {4}}\\\end{aligned}}}$

and therefore,

${\displaystyle {\begin{aligned}r_{2}(n)&=\pi \left({\frac {c_{1}(n)}{1}}-{\frac {c_{3}(n)}{3}}+{\frac {c_{5}(n)}{5}}-{\frac {c_{7}(n)}{7}}+{\frac {c_{11}(n)}{11}}-{\frac {c_{13}(n)}{13}}+{\frac {c_{15}(n)}{15}}-{\frac {c_{17}(n)}{17}}+\cdots \right)\\[6pt]r_{4}(n)&=\pi ^{2}n\left({\frac {c_{1}(n)}{1}}-{\frac {c_{4}(n)}{4}}+{\frac {c_{3}(n)}{9}}-{\frac {c_{8}(n)}{16}}+{\frac {c_{5}(n)}{25}}-{\frac {c_{12}(n)}{36}}+{\frac {c_{7}(n)}{49}}-{\frac {c_{16}(n)}{64}}+\cdots \right)\\[6pt]r_{6}(n)&={\frac {\pi ^{3}n^{2}}{2}}\left({\frac {c_{1}(n)}{1}}-{\frac {c_{4}(n)}{8}}-{\frac {c_{3}(n)}{27}}-{\frac {c_{8}(n)}{64}}+{\frac {c_{5}(n)}{125}}-{\frac {c_{12}(n)}{216}}-{\frac {c_{7}(n)}{343}}-{\frac {c_{16}(n)}{512}}+\cdots \right)\\[6pt]r_{8}(n)&={\frac {\pi ^{4}n^{3}}{6}}\left({\frac {c_{1}(n)}{1}}+{\frac {c_{4}(n)}{16}}+{\frac {c_{3}(n)}{81}}+{\frac {c_{8}(n)}{256}}+{\frac {c_{5}(n)}{625}}+{\frac {c_{12}(n)}{1296}}+{\frac {c_{7}(n)}{2401}}+{\frac {c_{16}(n)}{4096}}+\cdots \right)\end{aligned}}}$

r′_2s(n) (sums of triangles)

${\displaystyle r'_{2s}(n)}$ is the number of ways n can be represented as the sum of 2s triangular numbers (i.e. the numbers 1, 3 = 1 + 2, 6 = 1 + 2 + 3, 10 = 1 + 2 + 3 + 4, 15, ...; the n-th triangular number is given by the formula n(n + 1)/2.)

The analysis here is similar to that for squares. Ramanujan refers to the same paper as he did for the squares, where he showed that there is a function ${\displaystyle \delta '_{2s}(n)}$ such that ${\displaystyle r'_{2s}(n)=\delta '_{2s}(n)}$ for s = 1, 2, 3, and 4, and that for s > 4, ${\displaystyle \delta '_{2s}(n)}$ is a good approximation to ${\displaystyle r'_{2s}(n).}$

Again, s = 1 requires a special formula:

${\displaystyle \delta '_{2}(n)={\frac {\pi }{4}}\left({\frac {c_{1}(4n+1)}{1}}-{\frac {c_{3}(4n+1)}{3}}+{\frac {c_{5}(4n+1)}{5}}-{\frac {c_{7}(4n+1)}{7}}+\cdots \right).}$

If s is a multiple of 4,

${\displaystyle {\begin{aligned}\delta '_{2s}(n)&={\frac {({\frac {\pi }{2}})^{s}}{(s-1)!}}\left(n+{\frac {s}{4}}\right)^{s-1}\left({\frac {c_{1}(n+{\frac {s}{4}})}{1^{s}}}+{\frac {c_{3}(n+{\frac {s}{4}})}{3^{s}}}+{\frac {c_{5}(n+{\frac {s}{4}})}{5^{s}}}+\cdots \right)&&s\equiv 0{\pmod {4}}\\[6pt]\delta '_{2s}(n)&={\frac {({\frac {\pi }{2}})^{s}}{(s-1)!}}\left(n+{\frac {s}{4}}\right)^{s-1}\left({\frac {c_{1}(2n+{\frac {s}{2}})}{1^{s}}}+{\frac {c_{3}(2n+{\frac {s}{2}})}{3^{s}}}+{\frac {c_{5}(2n+{\frac {s}{2}})}{5^{s}}}+\cdots \right)&&s\equiv 2{\pmod {4}}\\[6pt]\delta '_{2s}(n)&={\frac {({\frac {\pi }{2}})^{s}}{(s-1)!}}\left(n+{\frac {s}{4}}\right)^{s-1}\left({\frac {c_{1}(4n+s)}{1^{s}}}-{\frac {c_{3}(4n+s)}{3^{s}}}+{\frac {c_{5}(4n+s)}{5^{s}}}-\cdots \right)&&s\equiv 1{\pmod {2}}{\text{ and }}s>1\end{aligned}}}$

Therefore,

${\displaystyle {\begin{aligned}r'_{2}(n)&={\frac {\pi }{4}}\left({\frac {c_{1}(4n+1)}{1}}-{\frac {c_{3}(4n+1)}{3}}+{\frac {c_{5}(4n+1)}{5}}-{\frac {c_{7}(4n+1)}{7}}+\cdots \right)\\[6pt]r'_{4}(n)&=\left({\frac {\pi }{2}}\right)^{2}\left(n+{\frac {1}{2}}\right)\left({\frac {c_{1}(2n+1)}{1}}+{\frac {c_{3}(2n+1)}{9}}+{\frac {c_{5}(2n+1)}{25}}+\cdots \right)\\[6pt]r'_{6}(n)&={\frac {({\frac {\pi }{2}})^{3}}{2}}\left(n+{\frac {3}{4}}\right)^{2}\left({\frac {c_{1}(4n+3)}{1}}-{\frac {c_{3}(4n+3)}{27}}+{\frac {c_{5}(4n+3)}{125}}-\cdots \right)\\[6pt]r'_{8}(n)&={\frac {({\frac {\pi }{2}})^{4}}{6}}(n+1)^{3}\left({\frac {c_{1}(n+1)}{1}}+{\frac {c_{3}(n+1)}{81}}+{\frac {c_{5}(n+1)}{625}}+\cdots \right)\end{aligned}}}$

Sums

Let

${\displaystyle {\begin{aligned}T_{q}(n)&=c_{q}(1)+c_{q}(2)+\cdots +c_{q}(n)\\U_{q}(n)&=T_{q}(n)+{\tfrac {1}{2}}\phi (q)\end{aligned}}}$

Then for s > 1,

${\displaystyle {\begin{aligned}\sigma _{-s}(1)+\cdots +\sigma _{-s}(n)&=\zeta (s+1)\left(n+{\frac {T_{2}(n)}{2^{s+1}}}+{\frac {T_{3}(n)}{3^{s+1}}}+{\frac {T_{4}(n)}{4^{s+1}}}+\cdots \right)\\&=\zeta (s+1)\left(n+{\tfrac {1}{2}}+{\frac {U_{2}(n)}{2^{s+1}}}+{\frac {U_{3}(n)}{3^{s+1}}}+{\frac {U_{4}(n)}{4^{s+1}}}+\cdots \right)-{\tfrac {1}{2}}\zeta (s)\\d(1)+\cdots +d(n)&=-{\frac {T_{2}(n)\log 2}{2}}-{\frac {T_{3}(n)\log 3}{3}}-{\frac {T_{4}(n)\log 4}{4}}-\cdots \\d(1)\log 1+\cdots +d(n)\log n&=-{\frac {T_{2}(n)(2\gamma \log 2-\log ^{2}2)}{2}}-{\frac {T_{3}(n)(2\gamma \log 3-\log ^{2}3)}{3}}-{\frac {T_{4}(n)(2\gamma \log 4-\log ^{2}4)}{4}}-\cdots \\r_{2}(1)+\cdots +r_{2}(n)&=\pi \left(n-{\frac {T_{3}(n)}{3}}+{\frac {T_{5}(n)}{5}}-{\frac {T_{7}(n)}{7}}+\cdots \right)\end{aligned}}}$